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leetcode solution Solution 2: Using ASCII Code Follow Up: If you want to return the longest substring not the length of it

admin 1月 08, 2021 0

https://leetcode.com/problems/longest-substring-without-repeating-characters/

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for “abcabcbb” is “abc”, which the length is 3. For “bbbbb” the longest substring is “b”, with the length of 1.

public class  {
    public int lengthOfLongestSubstring(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        Set<Character> set = new HashSet<Character>();
        int max = 0, i = 0, j = 0;
        while(j < s.length()){
            if(!set.contains(s.charAt(j))){
                set.add(s.charAt(j++));
                max = Math.max(max, set.size());
            } else {
                set.remove(s.charAt(i++));
            }
        }
        return max;
    }
}

Solution 2: Using ASCII Code

public class  {
    public int lengthOfLongestSubstring(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        boolean[] isUsed = new boolean[256];
        int max = 0, i = 0, j = 0;
        while(j < s.length()){
            if(!isUsed[s.charAt(j)]){
                max = Math.max(max, j - i + 1);
                isUsed[s.charAt(j++)] = true;
            } else {
                isUsed[s.charAt(i++)] = false;
            }
        }
        return max;
    }
}

Follow Up: If you want to return the longest substring not the length of it

public class  {
    public String lengthOfLongestSubstring(String s) {
        if(s == null || s.length() == 0){
            return null;
        }
        Set<Character> set = new HashSet<Character>();
        int max = 0, i = 0, j = 0, pos = 0;
        while(j < s.length()){
            if(!set.contains(s.charAt(j))){
                set.add(s.charAt(j++));
                if (max < set.size()) {
                    max = set.size();
                    pos = j - 1;
                }
            }
            else{
                set.remove(s.charAt(i++));
            }
        }
        return s.substring(pos + 1 - max, pos + 1);
    }
}